Reputation: 5183
How to convert hexadecimal string to single precision floating point in Java?
For example, how to implement:
float f = HexStringToFloat("BF800000"); // f should now contain -1.0
I ask this because I have tried:
float f = (float)(-1.0);
String s = String.format("%08x", Float.floatToRawIntBits(f));
f = Float.intBitsToFloat(Integer.valueOf(s,16).intValue());
But I get the following exception:
java.lang.NumberFormatException: For input string: "bf800000"
Upvotes: 10
Views: 24322
Reputation: 24748
public class Test {
public static void main (String[] args) {
String myString = "BF800000";
Long i = Long.parseLong(myString, 16);
Float f = Float.intBitsToFloat(i.intValue());
System.out.println(f);
System.out.println(Integer.toHexString(Float.floatToIntBits(f)));
}
}
Upvotes: 26
Reputation: 29240
You need to convert the hex value to an int (left as an exercise) and then use Float.intBitsToFloat(int)
Upvotes: 2